Question

You have a 1 mg of fluorochrome X with a molecular weight 515.0g that can be dissolved in DMSO at a
concentration up to 5 mM.

You want to stain your cells at a final concentration 5 uM at a staining volume 100 ul.

Q: Outline just how exactly you will dissolve the powder (volumes) to make stocks, what diluent, aliquots, what you will do with left overs (how to store
and/or dispose them) and what volumes you will add to the cells to achieve final concentration.

Answer 1

Given the molecular weight of fluorochrome X = 515 g / mole.

The solvent is DMSO and we need to prepare a stock so that the stain is 5 uM in a total volume of 100 uL. We could prepare a 10X stock of 50 uM so that we get to add 1 / 10th of the stain to get a final concentration of 50 / 10 = 5 uM.

Molarity = moles / liter

A 1 M solution of fluorochrome X = 515 g / L = 515 mg / mL.

The maximum concentration of fluorochrome X that is soluble in DMSO is 5mM.

For a 5 mM or a 0.005 M stock , we would need : 515 * 0.005 = 2.6 mg / mL. We would dissolve 2.6 mg in 1 mL DMSO and use this as the primary stock.

From this primary stock we would prepare a working stock of 50 uM or 0.05 mM :

We would calculate the dilution needed to get 50 uM from 5 mM or 5000 uM stock solution:

50 / 5000 = 5 / 500 = 1 / 100.

Therefore, we need to make a 10^-2 or a 1 / 100 dilution. We would take 10 uL of the primary stock and add 990 uL diluent to get 1 mL of 50 uM working stock.

We would use 10 uL of the above stock to stain cells in a total volume of 100 uL (cells in 90 uL solution + 10 uL 50 uM Fluorochrome X)